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There is a well-known alphametic story where a poor college student sends a telegram with the words "SEND MORE MONEY" to his parents, asking for more money, and asking to fill in each letter with a different digit in order to figure out how much he is asking for. In another alphametic story the same poor college student sends a telegram with the words "SEND A GIFT":

     SEND = A * GIFT

The parents send him a gift. Although the gift is odd and what the parents send is odd and everything related is odd, the poor college student is happy with it.

Which digit does each letter represent? (Please present the full analysis how these digits can be determined. Every letter stands for a digit in base-10 representation, different letters stand for different digits, and leading digits are always non-zero.)

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Solution:

There are 15 possible solutions:
$4827 = 3 * 1609$
$5427 = 3 * 1809$
$8463 = 7 * 1209$
$9706 = 2 * 4853$
$3690 = 2 * 1845$
$6970 = 2 * 3485$
$7690 = 2 * 3845$
$9630 = 2 * 4815$
$9670 = 2 * 4835$
$9730 = 2 * 4865$
$9370 = 2 * 4685$
$7852 = 4 * 1963$
$7236 = 4 * 1809$
$8316 = 4 * 2079$
$7854 = 6 * 1309$

Starting point

"The gift is odd". This is the part.
It means the amount is an odd number. So we conclude T has to be odd digit.
$A$ and $T$ obviously can not be $1$
Since $A >=2$ then $G <=4$
Also A cannot be 5 since it will result in $D = 5$
Also A cannot be 9 because it will result in $S = 9$ or go over 4 digits in the result
So A can be one of [2,3,4,6,7,8]

[EDIT]

As Trenin pointed out A can also be even. In my original answer I handled A as odd, that's why the cases are handled in this way..A odd then A even.

now start filling in the blanks using the values we already know

Case 1:

A = 3, T = 7. We immediately see that $G =2$ (otherwise we get to $S = 9$ or go over 4 digits).
The carriage from $3*I + carriage from 3*f + 2$ should be one of [0,2,3] (otherwise we get S = 7 or we go over 4 digits). But 3 cannot be achieved since we cannot get the carriage from 3*f to be 3.
The carriage can be 0 only if $I = 0$ (it cannot be 2 or 3 because they are already taken). We have in this case
20F7 * 3 = 6EN1 F cannot be 4 because it will result in N being 4, F cannot be 5 or 9 since it will result in N being 7 (already taken), F cannot be 8 since it will result in E = 2 (already taken).
So no valid solution here.
The carriage can be 2 only if I is [6,8,9].
Trying every combination for I = 6 and F any available digit gets us to conflicts every time.

Conclusion $A = 3, T = 7$ - not possible

Case 2.

A = 3, T = 9.
We get G is one of [1,2]
Considering G = 1.
1IF9 * 3 = SEN7.
Considering F = 0 we get 1I09 * 3 = SE27.
Possible values for I are [4,5,6,8].
Doing the math for all possible values we get 2 matches. I one of [6,8]
This results in 2 possible solutions:
$4827 = 3 *1609$ and $5427 = 3 * 1809$

For G = 2 we end up with F being one of [6,8].
For F = 6 there is no possible value for I. (we end up with an already used value for E).
For F = 8 There is no possible value for I.
Conclusion: No other possible values that what we found above

Case 3:

A = 7, T = 3. This results in G = 1.
Starting with F = 0 we get no valid value for I.
For F = 2 again no valid value for I.
Going through all the possible values of F ([4,5,6,8,9]) we get the same result. Nothing valid for I.

Conclusion: this doesn't work

Case 4:

A = 7, T = 9. We see immediately that G = 1.
Trying with F one of [0,2,3,4,5,6,8] and following the same process as for the other cases (got lazy here so I won't list all the possible outcomes) we end up with the one solution for F = 0.

$8463 = 7 * 1209$

[EDIT] After Trenin's comment.

In case A and SEND can be even numbers, in addition to the cases above we get:

Case 5.1:

A = 2, T = 3 => G can be only 4 (for 1,2,3 we get duplicate digits).
in this case F can be one of [5,8,9].
IN this case we get only one valid value $9706 = 2 * $4853$

Case 5.2:

A =2, T = 5 => G can be 1, 3 or 4. otherwise we get conflicts in the digits.
Starting with G = 1. We get the possible values for F [3,4,6,7,8,9].
Calculating step by step we end up with these solutions:
$3690 = 2 * 1845$
continuing with G = 3 we get the possible values for F [4,6,8]
Doing the math for all of them we get: $6970 = 2 * 3485$
$7690 = 2 * 3845$
for G = 4 we get the possible values for F [1,3,6,8].
Doing the math we get $9630 = 2 * 4815$
$9670 = 2 * 4835$
$9730 = 2 * 4865$
$9370 = 2 * 4685$

Case 5.3

A = 2, T = 7. Results in G being 1 or 3.
Doing the same steps as above, calculating the possible values for F and then filling in the missing value for I we get no valid results.

Case 5.4

A = 4, T = 3. Results in G being 1 or 2.
Doing the same steps as above, calculating the possible values for F and filling in the missing value for I we get one valid result:
$7852 = 4*1963$

Case 5.5 to 5.15 (got tired of typing )

A = 4, T = 5 gives no valid result
A = 4, T = 7 no results again
A = 4, T = 9 gives out
$7236 = 4 * 1809$
$8316 = 4 * 2079$
Starting A = 6 and further is even easier since G can be only 1.
A = 6, T = 3 => nothing
A = 6, T = 5 => nothing
A = 6, T = 7 => nothing A = 6, T = 9 we get:
$7854 = 6 * 1309$
A = 8, T = 3 => nothing
A = 8, T = 5 => nothing
A = 8, T = 7 => nothing
A = 8, T = 5 => nothing

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  • $\begingroup$ Why does $A$ have to be odd? The clue is only that the gift is odd. If the clue was that the sending of the gift was odd, then I would conclude $SEND$ is odd and agree that both $T,A,D$ are all odd. But from the clue, I don't think you can infer $A$ is odd, but only $GIFT$ and thus only $T$ is odd. $\endgroup$ – Trenin Mar 29 '16 at 12:27
  • $\begingroup$ @Trenin. Hmmm...you might be right on this. I saw the word odd and got all fired up. Anyway, the solutions I found are still valid. I will start digging for the other solutions as well. Thanks for this. $\endgroup$ – Marius Mar 29 '16 at 12:30
  • $\begingroup$ @Trenin. Fixed it. Thanks again for the tip. I would split the points with you on this if I could :). $\endgroup$ – Marius Mar 29 '16 at 13:14

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